Question

The ionization energy of the electron in the hydrogen atom in its ground state is $136eV$. The atoms are excited to higher energy levels too emits radiation of $6$ wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between:

• A. $n=3$ to $n=2$ states
• B. $n=3$ to $n=1$ states
• C. $n=2$ to $n=1$ states
• D. $n=4$ to $n=3$ states

Answer 1

The correct option is D $n=4$ to $n=3$ states

Number of spectral lines obtained due to transition of electron from $n^{th}$ orbit to lower orbit $N=\frac{n(n-1)}{2}$

And for maximum wavelength, the difference between the orbits of the series should be minimum.

Number of spectral lines $N=\frac{n(n-1)}{2}=6$

$\Rightarrow \frac{n(n-1)}{2}=6$

$\Rightarrow n^2-n-12=0$

$\Rightarrow (n-4)(n+3)=0$

$n=4$

Now as the first line of the series has the maximum wavelength, therefore, electrons jump from the $4^{th}$ orbit to $3^{rd}$ orbit.